d = 2R\sin\frac{\varphi}{2}<r \Longleftrightarrow
\sin\frac{\varphi}{2}<\frac{r}{2R}
\Longleftrightarrow\frac{\varphi}{2}<\arcsin\frac{r}{2R}\Longleftrightarrow
\varphi<2\arcsin\frac{r}{2R}
\\|\varphi_{1}-\varphi_{2}|<2\arcsin\frac{r}{2R}, \quad
|\varphi_{1}-\varphi_{2}\pm 2\pi|<2\arcsin\frac{r}{2R},\ {a}
\quad\Omega=[0,2\pi]\times [0,2\pi] \\
P=\frac{{4\arcsin\frac{r}{2R}\cdot2\pi}}{\left({2\pi}\right)^2}=
\frac{2}{\pi}\arcsin\frac{r}{2R}